Resolução Marion [Mec.Clássica] - Capítulo 10

10-1. The accelerations which we feel at the surface of the Earth are the following:

(1) Gravitational : 2980 cm/sec (2) Due to the Earth’s rotation on its own axis:

r πω −

10-2. The fixed frame is the ground. y a θ x

The rotating frame has the origin at the center of the tire and is the frame in which the tire is at rest.

From Eqs. (10.24), (10.25):

334 CHAPTER 10 Now we have

cos sin 0f ra r

Va r

Ri ri v a

Substituting gives 2 cos sinf v a a r cos sin 1f v

We want to maximize fa, or alternatively, we maximize 2 fa:

2 cos cos 2 sin sin

2 2c os sin va v a a va v a r aθ θθθθ

2 cos 2 cos

0 when tan fd av a

(Taking a second derivative shows this point to be a maximum.) n implies cos ar v v ar v θθ==+ta and sin ar

Substituting into (1) arva v a

r ar v a r v ai

This may be written as

MOTION IN A NONINERTIAL REFERENCE FRAME 335

This is the maximum acceleration. The point which experiences this acceleration is at A:

tanθ=

The only forces acting are centrifugal and friction, thus 2smgmrµω=, or

2 sgrµω=

10-4. Given an initial position of (–0.5R,0) the initial velocity (0,0.5ωR) will make the puck motionless in the fixed system. In the rotating system, the puck will appear to travel clockwise in a circle of radius 0.5R. Although a numerical calculation of the trajectory in the rotating system is a great aid in understanding the problem, we will forgo such a solution here.

⋅, since we are given in the problem that the initial velocity is at an angle of 45° to the x-axis. We will vary over some range that we know satisfies the condition that the path cross over . We can start by looking at Figures 10-4e and 10-4f, which indicate that we want . Trial and error can find a trajectory that does loop but doesn’t cross its path at all, such as

eyeball-suitable. This may be an entirely satisfactory answer, depending on the inclinations of the instructor. An interpolation over several trajectories would show that an accurate answer to the problem is , which exits the merry-go-round at 3.746 s. The figure shows this solution, which was numerically integrated with 200 steps over the time interval.

x (m)

10-6. z m r

Consider a small mass m on the surface of the water. From Eq. (10.25)

In the rotating frame, the mass is at rest; thus, eff0=F. The force F will consist of gravity and the force due to the pressure gradient, which is normal to the surface in equilibrium. Since

mω r θ′

Since F, the sum of the gravitational and centrifugal forces must also be normal to the surface. eff 0=

Thus θ′ = θ.

tan tan rg ωθθ==′

MOTION IN A NONINERTIAL REFERENCE FRAME 337 but tan dzdr θ=

Thus

2 constant 2

The shape is a circular paraboloid.

zr g ω=+

10-7. For a spherical Earth, the difference in the gravitational field strength between the poles and the equator is only the centrifugal term:

2 polesequatorggRω−=

For and R = 6370 km, this difference is only 3457.310radsω−=×⋅1−2 mms−⋅. The disagreement with the true result can be explained by the fact that the Earth is really an oblate spheroid, another consequence of rotation. To qualitatively describe this effect, approximate the real Earth as a somewhat smaller sphere with a massive belt about the equator. It can be shown with more detailed analysis that the belt pulls inward at the poles more than it does at the equator. The next level of analysis for the undaunted is the “quadrupole” correction to the gravitational potential of the Earth, which is beyond the scope of the text.

10-8.

y z

Choose the coordinates x, y, z as in the diagram. Then, the velocity of the particle and the rotation frequency of the Earth are expressed as

cos , 0, sinz ω λω λ

This acceleration is directed along the y axis. Hence, as the particle moves along the z axis, it will be accelerated along the y axis: